A train covers a distance of 600 km at x km/hr. Had the speed been (x + 20) km/hr, the time taken to cover the same distance would have been reduced by 5 hours. Write down an equation in x and solve it to evaluate x.

By formula,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Time taken by train to cover distance of 600 km at x km/hr = $\dfrac{600}{x}$hrs

Time taken by train to cover distance of 600 km at (x + 20) km/hr = $\dfrac{600}{x + 20}$hrs

Given,

On increasing the speed, the time taken is reduced by 5 hours.

$\Rightarrow \dfrac{600}{x} - \dfrac{600}{x + 20} = 5 \\[1em] \Rightarrow \dfrac{600(x + 20) - 600x}{x(x + 20)} = 5 \\[1em] \Rightarrow \dfrac{600x + 12000 - 600x}{x^2 + 20x} = 5 \\[1em] \Rightarrow 12000 = 5(x^2 + 20x) \\[1em] \Rightarrow 12000 = 5x^2 + 100x \\[1em] \Rightarrow 0 = 5x^2 + 100x - 12000 \\[1em] \Rightarrow 5x^2 + 100x - 12000 = 0 \\[1em] \Rightarrow 5(x^2 + 20x - 2400) = 0 \\[1em] \Rightarrow x^2 + 20x - 2400 = 0 \\[1em] \Rightarrow x^2 + 60x - 40x - 2400 = 0 \\[1em] \Rightarrow x(x + 60) - 40(x + 60) = 0 \\[1em] \Rightarrow (x - 40)(x + 60) = 0 \\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 60) = 0 \text{….[Using zero-product rule]} \\[1em] \Rightarrow x = 40 \text{ or } x = -60$

Since speed cannot be negative: x = 40km/hr

Hence, obtained equation is x² + 20x - 2400 = 0 and x = 40 km/hr.